Problem: What is the value of the following logarithm? $\log_{8} \left(\dfrac{1}{64}\right)$
Explanation: If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $8^{y} = \dfrac{1}{64}$ In this case, $8^{-2} = \dfrac{1}{64}$, so $\log_{8} \left(\dfrac{1}{64}\right) = -2$.